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Dmitry [639]
3 years ago
8

In parallelogram abcd, m

Mathematics
2 answers:
just olya [345]3 years ago
6 0
What aboout it? please clarify.
ExtremeBDS [4]3 years ago
5 0
What's the rest of the problem? Can't help if we don't have a picture of some sort or if we don't have the rest of the question
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HELP I WILL GIVE YOU BRAINLIEST <br><br> Question: What is the solution to this quadratic?
Nataly_w [17]
(5,0) if it’s asking the solution or do you want the equation? Technically the solution would be where the parabola touched the x axis
7 0
3 years ago
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Mary is spending money at the average rate of $5 per day. After 14 days she has $68 left. The amount left depends on the number
Radda [10]

Answer:

A. Y-68=-5(x-14)

Step-by-step explanation:

Mary is spending money at the average rate of $5 per day. After 14 days she has $68 left. The amount left depends on the number of days that have passed. write an equation for the situation

We find the equation using the point slope equation of a line. This is given as:

y - y1 = m(x-x1)

The slope intercept form will be

y = -5x + b

since the amount decreases by $5 per day...

Our slope is -5 so m=-5

We have a point (x1,y1) = (14, 68)

Hence, we have:

y-68 = -5(x-14)

Option A is correct

4 0
3 years ago
Is 2/5 and 8/20 equivalent
dolphi86 [110]
Yes!
explanation:
8 divided by 4 equals 2 and 20 divided by 4 is 5
4 0
3 years ago
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Expanded form for 41.32
Anna35 [415]
It is:

40+1+0.3+0.02=41.32



4 0
3 years ago
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Find the derivative of the function at P 0 in the direction of A. ​f(x,y,z) = 3 e^x cos(yz)​, P0 (0, 0, 0), A = - i + 2 j + 3k
Alik [6]

The derivative of f(x,y,z) at a point p_0=(x_0,y_0,z_0) in the direction of a vector \vec a=a_x\,\vec\imath+a_y\,\vec\jmath+a_z\,\vec k is

\nabla f(x_0,y_0,z_0)\cdot\dfrac{\vec a}{\|\vec a\|}

We have

f(x,y,z)=3e^x\cos(yz)\implies\nabla f(x,y,z)=3e^x\cos(yz)\,\vec\imath-3ze^x\sin(yz)\,\vec\jmath-3ye^x\sin(yz)\,\vec k

and

\vec a=-\vec\imath+2\,\vec\jmath+3\,\vec k\implies\|\vec a\|=\sqrt{(-1)^2+2^2+3^2}=\sqrt{14}

Then the derivative at p_0 in the direction of \vec a is

3\,\vec\imath\cdot\dfrac{-\vec\imath+2\,\vec\jmath+3\,\vec k}{\sqrt{14}}=-\dfrac3{\sqrt{14}}

3 0
3 years ago
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